CosmicBills Posted December 20, 2004 Share Posted December 20, 2004 I am afraid of math, but was wondering this: If each NFL game is a 50/50 Win Loss proposition, what is the statistical probibility that Denver, Baltimore and Jax lose at least one of their final two games? or, in other words... It's like 4 people flipping a coin twice. What's the chances that all four people get a Heads at least once? Can anyone solve that equation? I would love to know. Thanks! Link to comment Share on other sites More sharing options...
daquixers Posted December 20, 2004 Share Posted December 20, 2004 I failed Geometry --- Link to comment Share on other sites More sharing options...
Guest Guest Posted December 20, 2004 Share Posted December 20, 2004 Simple 50/50. The events are independent of each other. If the probablility is 50/50 for each game, it is 50/50 for the scenario you describe. What are they teaching you kids these days in school ... how to roll a condom on? Link to comment Share on other sites More sharing options...
daquixers Posted December 20, 2004 Share Posted December 20, 2004 Simple 50/50. The events are independent of each other. If the probablility is 50/50 for each game, it is 50/50 for the scenario you describe. What are they teaching you kids these days in school ... how to roll a condom on? 168175[/snapback] You roll them on ?? Link to comment Share on other sites More sharing options...
JimBob2232 Posted December 20, 2004 Share Posted December 20, 2004 Its 1/8th. If you flip a coin 3 times, the likelyhood it will turn up heads (or tails) 3 times in a row is 1 in 8 Link to comment Share on other sites More sharing options...
CosmicBills Posted December 20, 2004 Author Share Posted December 20, 2004 Simple 50/50. The events are independent of each other. If the probablility is 50/50 for each game, it is 50/50 for the scenario you describe. What are they teaching you kids these days in school ... how to roll a condom on? 168175[/snapback] That can't be the answer...can it? Because wouldn't you have to factor in the chances that only 2 of the 3 teams get a Heads, or the chance that only 1 of the 3 teams get a Heads... It's not 50/50. Link to comment Share on other sites More sharing options...
CosmicBills Posted December 20, 2004 Author Share Posted December 20, 2004 Its 1/8th. If you flip a coin 3 times, the likelyhood it will turn up heads (or tails) 3 times in a row is 1 in 8 168178[/snapback] Ah, thank you:) So we have a 12% chance... Looks grim when you see it in that format, but it's better than nothing! Link to comment Share on other sites More sharing options...
SDS Posted December 20, 2004 Share Posted December 20, 2004 Simple 50/50. The events are independent of each other. If the probablility is 50/50 for each game, it is 50/50 for the scenario you describe. What are they teaching you kids these days in school ... how to roll a condom on? 168175[/snapback] I don't know the right answer, but I know that you're wrong.... flipping a penny is an independent event too, but the prob. that it is flipped tails 3x in a row is not 50%. Link to comment Share on other sites More sharing options...
JimBob2232 Posted December 20, 2004 Share Posted December 20, 2004 Ah, thank you:) So we have a 12% chance... Looks grim when you see it in that format, but it's better than nothing! And its sure alot better than I would have thought we had back in october... Link to comment Share on other sites More sharing options...
CosmicBills Posted December 20, 2004 Author Share Posted December 20, 2004 And its sure alot better than I would have thought we had back in october... 168189[/snapback] Amen! GO BILLS! Link to comment Share on other sites More sharing options...
SDS Posted December 20, 2004 Share Posted December 20, 2004 Ah, thank you:) So we have a 12% chance... Looks grim when you see it in that format, but it's better than nothing! 168182[/snapback] I doubt that is correct either because you only need one loss out of 2 possible tries for each team. If it was the final game - yes, but not now. I think it is 0.75^3 = 42% WW, LW, WL, LL = 75% success for each team.... Link to comment Share on other sites More sharing options...
CosmicBills Posted December 20, 2004 Author Share Posted December 20, 2004 I doubt that is correct either because you only need one loss out of 2 possible tries for each team. If it wa sthe final game - yes, but not now. 168193[/snapback] Oh, great point! See, this is why math and I don't get along too well. Link to comment Share on other sites More sharing options...
SDS Posted December 20, 2004 Share Posted December 20, 2004 Oh, great point! See, this is why math and I don't get along too well. 168196[/snapback] I think I gave the right answer in my edited post. Link to comment Share on other sites More sharing options...
crackur Posted December 20, 2004 Share Posted December 20, 2004 GOOD DEAL SDS Link to comment Share on other sites More sharing options...
CosmicBills Posted December 20, 2004 Author Share Posted December 20, 2004 I think I gave the right answer in my edited post. 168199[/snapback] Thanks so much. I know it's meaningless, and even though I hate math, I like numbers. I am a walking contradiction I guess. Link to comment Share on other sites More sharing options...
JimBob2232 Posted December 20, 2004 Share Posted December 20, 2004 I have posed the question to some people who know much more than I do regarding statistics...I will post the results as I get them.... Link to comment Share on other sites More sharing options...
SDS Posted December 20, 2004 Share Posted December 20, 2004 Thanks so much. I know it's meaningless, and even though I hate math, I like numbers. I am a walking contradiction I guess. 168202[/snapback] not to bum you out, but assuming I'm right - I believe I am - then if you include OUR odds of WW then all the scenarios play out to a 10.5% chance of occuring... Link to comment Share on other sites More sharing options...
Guest TSquareBoston Posted December 20, 2004 Share Posted December 20, 2004 42.2% Assuming equal (50% probability) in each game: Each independent team has a 75% chance of losing at least one game: 4 possible combinations, of which 3 consist of a loss: W/W,W/L,L/W and L/L. 3/4=75% The chance of all three having a positive outcome (at least one loss) is 75% x 75% x 75%. Dated a math major way back. Link to comment Share on other sites More sharing options...
ATBNG Posted December 20, 2004 Share Posted December 20, 2004 Well, I don't think that you can make a reasonable analysis assuming every game is a 50/50 proposition. We know that isn't the case - you'd be hard pressed to find someone make you an even money bet on next week's San Fran/Buffalo game. In this case, you have to factor in the lines on the games. Here's my projections: Buff at SF - Buffalo by 13 Pitt at Buff - Pitt 3 (if 1 seed isn't wrapped up) Buff 1-3 (if Pitt wraps up one seed) Denver at Tennessee - Denver -3 Indy at Denver - Denver -3 (if Indy can't improve their position, which is very likely...if this game were in week 16, Indy would be a three point favorite - this is really a killer for the Bills) Houston at Jacksonville - Jags -5.5 Jacksonville at Oakland - Jacksonville -6 (if they can win to get in - a scenario that will inflate the line....normally that's a Jacks -3) Baltimore at Pittsburgh - Pittsburgh -7 Miami at Baltimore - Baltimore -12 There's also the Jets scenario.... Pats at Jets - New England -3 Jets at St. Louis - Jets -8 Since Denver and Jacksonville are favored in both their games, there's a higher than 25% probability (nominal coin flipping odds) that either one will win both. I'd put the Jags somewhere around 50% and the Broncos at 35%. I'd put Baltimore at around 15%. The Bills are highly dependent on the value of the game to Pittsburgh. I think the Bills have a 20% shot or so at getting into this thing. Link to comment Share on other sites More sharing options...
MDH Posted December 20, 2004 Share Posted December 20, 2004 Simple 50/50. The events are independent of each other. If the probablility is 50/50 for each game, it is 50/50 for the scenario you describe. What are they teaching you kids these days in school ... how to roll a condom on? 168175[/snapback] You have to love someone insulting other's education when the answer they come to isn't even close to being the correct answer. Try taking a Stats class before coming back to sit on your high horse. Link to comment Share on other sites More sharing options...
Mikie2times Posted December 20, 2004 Share Posted December 20, 2004 I am afraid of math, but was wondering this: If each NFL game is a 50/50 Win Loss proposition, what is the statistical probibility that Denver, Baltimore and Jax lose at least one of their final two games? or, in other words... It's like 4 people flipping a coin twice. What's the chances that all four people get a Heads at least once? Can anyone solve that equation? I would love to know. Thanks! 168163[/snapback] I just made a post about this with my own calculations. Here is a simple way to figure out the odds of those teams winning out. Take your odds for each game. For example in my post I have the Jaguars at 6/10 vs Houston. In other words for every 10 games they play against Houston in Jacksonville they will win 6 of them. Then you take the odds of your second game, in Jacksonville's case I have them at 8/10 vs Oakland. Turn those fractions into percentages and multiply them together. So its . 60*.80= 48%. So within my estimates the Jaguars have a 48% chance at winning out. Link to comment Share on other sites More sharing options...
eSJayDee Posted December 20, 2004 Share Posted December 20, 2004 f each NFL game is a 50/50 Win Loss proposition, what is the statistical probibility that Denver, Baltimore and Jax lose at least one of their final two games? I think you asked 2 questions so, I'll answer this one. The easiest way to visualize this is to look at the chance it doesn't occur and subtract that from 100%. If a team fails to lose at least one game, that means they win both. The odds of that (independent) are 25%. So the odds of them losing at least 1 is 75%. You want to know the odds of all happening together, that's .75*.75*.75 =27/64=~42.8%. I was a math major, BTW (concentrating in calculus, though). Also, give SDS the gold star. He got the answer 1st. I figured it was easiest to figure the answer myself than to go over which previous posts were right & which ones were wrong. Link to comment Share on other sites More sharing options...
Guest cal t Posted December 20, 2004 Share Posted December 20, 2004 I am afraid of math, but was wondering this: If each NFL game is a 50/50 Win Loss proposition, what is the statistical probibility that Denver, Baltimore and Jax lose at least one of their final two games? or, in other words... It's like 4 people flipping a coin twice. What's the chances that all four people get a Heads at least once? Can anyone solve that equation? I would love to know. Thanks! 168163[/snapback] Let me correct the Math here.If each game is a 50/ 50 chance.The each individual team has a 25% chance of losing or winning both games. Each team has a 50% chance of winning one/losing one.If this is the case...you multiple .50 x .50 x .50 and you get .125 or 12.5 percent.-BUT--there is also a .25x.25 x.25 chance of allthese teams losing both games(if they dont play eachother) or about 2 percent(less actually) BUTthere is a chance that one team lose both & the others lose 1--.25 x .5 x .5=.625 or approx 6 percent. BUT--2 teams losing both plus one winning 1---.25x.25 x.5=3.125..So add 12.5 plus 1.5625plus6.25plus3.125...soooo There is a 23.4375 % chance that the Bills will make the playoffs ASSUMING 2 Bills wins--using the 50 percent scenario that you chose. Link to comment Share on other sites More sharing options...
Fan in San Diego Posted December 20, 2004 Share Posted December 20, 2004 Here is my entry in the stats derby 4 events of 50/50 2 * 2 * 2 *2 =16 1 in 16 or 6.25 % chance of 4 heads in a row. Link to comment Share on other sites More sharing options...
Guest cal t Posted December 20, 2004 Share Posted December 20, 2004 I think you asked 2 questions so, I'll answer this one.The easiest way to visualize this is to look at the chance it doesn't occur and subtract that from 100%. If a team fails to lose at least one game, that means they win both. The odds of that (independent) are 25%. So the odds of them losing at least 1 is 75%. You want to know the odds of all happening together, that's .75*.75*.75 =27/64=~42.8%. I was a math major, BTW (concentrating in calculus, though). Also, give SDS the gold star. He got the answer 1st. I figured it was easiest to figure the answer myself than to go over which previous posts were right & which ones were wrong. 168319[/snapback] Sorry you were both wrong. sincerely, cal t ...statistics expert. You have to get the odds of all possibilities putting the bills in the playoffs.Those scenarios are all 3 teams losing both. all 3 teams losing once, 1 losing both 2 winning 1,2 losing both 1 winning 1.then you add these......and you ASSUME the Bills win their 2(for true odds--you'd need to add that in--but we are ASSUMING 2 Bills wins)...and you get the 24 + % figure. Thank you, You've been a lovely audience. Link to comment Share on other sites More sharing options...
Recommended Posts